Problem: $\text E = \left[\begin{array}{rrr}5 & 5 & 1 \\ 4 & 0 & 3\end{array}\right]$ and $\text B = \left[\begin{array}{rr}-1 & 5 \\ 1 & 2 \\ -2 & 3\end{array}\right]$ Let $\text {H = EB}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{B}$. $ \text {H}=\left[\begin{array}{rr}{5} & {5} & {1} \\ 4 & 0 & 3\end{array}\right]\left[\begin{array}{rr} {-1} & 5 \\ {1} & 2 \\ {-2} & 3\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(5,5,1)\cdot(-1,1,-2)\\\\ &=5 \cdot -1+5\cdot 1 + 1\cdot -2\\\\ &=-2 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot 5+5\cdot 2 + 1\cdot 3 = 38$ (Choice B) B $4 \cdot -1+0\cdot 1 + 3\cdot -2 = -10$ (Choice C) C $5 \cdot 0 - 1\cdot 5 = -5$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-2 & 38 \\ -10 & 29\end{array}\right]$